✏️SLOPES
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[tex] \underline{\mathbb{PROBLEM}:} [/tex] Joshua says that the system x + 2y = 4 and x + 2y = 2 has no solution. Which of the following reasons would support his statement?
- I. The graph of the system of equations shows parallel lines.
- II. The graph of the system of equations shows intersecting lines.
- III. The two lines have the same slope but different y-intercepts.
- a. I and II
- b. I and III
- c. II and III
- d. I, II, and III
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[tex] \underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad\Large » \tt\: \green{B. \: I \: and \: III} [/tex]
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[tex] \underline{\mathbb{SOLUTION}:} [/tex] Determine the slope of the two linear equations by rearranging it into y-intercept form. In which [tex] m [/tex] is the slope of the line and [tex] b [/tex] is the y-intercept.
[tex] \begin{align} & \bold{Formula:} \\ & \boxed{y = mx + b} \end{align} [/tex]
- [tex] \begin{cases} x + 2y = 4 \\ x + 2y = 2 \end{cases} [/tex]
- [tex] \begin{cases} 2y = \text-x + 4 \\ 2y = \text-x + 2 \end{cases} [/tex]
- [tex] \begin{cases} \frac{\cancel2y}{\cancel2} = \frac{\text-x + 4}{2} \\ \frac{\cancel2y}{\cancel2} = \frac{\text-x + 2}{2} \end{cases} [/tex]
- [tex] \begin{cases} y = \text-\frac{1}{2}x + 2 \\ y = \text-\frac{1}{2}x + 1 \end{cases} [/tex]
» Now, we can see that the two linear equations have the same slope which is -½ but has different y-intercepts. Thus we can say that these lines are parallel.
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(ノ^_^)ノ
